\(\therefore\) Charge on one hydrogen atom

\(=q_{e}+q_{p}=-e+(e+\Delta e)=\Delta e\)

Since a hydrogen atom carry a net charge \(\Delta e\)

\(\therefore \quad\) Electrostatic force,

\(F_{e} \frac{1}{4 \pi \varepsilon_{o}} \frac{(\Delta e)^{2}}{d^{2}}.........(i)\)

will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as

\(F_{g}=\frac{G m_{h} m_{h}}{d^{2}}.........(ii)\)

since, the net force on the system is zero, \(F_{e}=F_{g}\) Using eqns. \((i)\) and \((ii)\), we get

\({\frac{(\Delta e)^{2}}{4 \pi \varepsilon_{o} d^{2}}=\frac{G m_{h}^{2}}{d^{2}}}\)

\({(\Delta e)^{2}=4 \pi \varepsilon_{o} G m_{h}^{2}}\)

\({=6.67 \times 10^{-11} \times\left(1.67 \times 10^{-27}\right)^{2} /\left(9 \times 10^{9}\right)}\)

\({\Delta e=10^{-37} \,\mathrm{C}}\)